sphere and Q is any point on the other sphere, wat is the maximum possible length of PQ?
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2006-09-21 22:15:22 · 13 answers · asked by smart_eluh 4 in Science & Mathematics ➔ Mathematics
a note to Aragosta n the others of th kind. am not doing any home work. am doing revision. these are some of the questions which I cldnt reason out how. thats why am asking for the working. u shld hav a look to several maths problems hav answered today and know that I av a passion for maths.
2006-09-21 23:01:56 · update #1
If the two spheres are tangent to each other, then P and Q are furthest apart when they are on opposite sides of the spheres from one another. PQ then goes through the center of both spheres, and is the sum of their diameters, 14 + 8 = 22.
2006-09-21 22:22:03 · answer #1 · answered by Anonymous · 0⤊ 0⤋
i am 15 years old by the way so if i get it wrong well sorry
well if the radius of a sphere is 7 then the diameter is 14 and if the radius of the other spere is 4 then the diameter is 8 if u put those to spheres are pput together then adding those diametere together the answer is 22
2006-09-21 22:25:31 · answer #2 · answered by !!David!! 2 · 0⤊ 0⤋
The sum of the 2 diameters. That's the longest length in a circle.
PQ=2(7)+2(4)=22
2006-09-21 22:21:37 · answer #3 · answered by Anonymous · 0⤊ 0⤋
If PQ is a tangent to the two spheres having radius 4 and 7 then the maximum length of PQ will be the distance between the centers of two circles
Note: Let O1 and O2 are centers of two circles. O1P and O2Q will be perpendicular to PQ .PO1O2 Q will be rectangle. PQ=O1O2= distance between two centers
2006-09-22 05:16:18 · answer #4 · answered by Amar Soni 7 · 0⤊ 0⤋
Assuming that by 'tangent to each other' means they touch each other, the max distance, PQ = PA + AQ = D1 + D2 = 2R1+2R2 =2x7 + 2x4 = 14 + 8 = 22. (No units are given in the question).
Here, A is the point of contact, the points P and Q are at the opposite end of the line joining the two centres C1 and C2 with A.
2006-09-21 22:27:32 · answer #5 · answered by Lizatom 1 · 0⤊ 0⤋
the max length will arise when the two points p and Q will be on the two ends of the line joining two centres of the spheres. so required length is the sum of the two diameters of the spheres and it is = 2*7+2*4=22 m
2006-09-22 00:58:15 · answer #6 · answered by Anonymous · 0⤊ 0⤋
If the small sphere is outside of the large one, then the answer is 22 (the sum of the two diameters). If the small sphere is *inside* the large one, the answer is 14 (the diameter of the large sphere using the point of contact).
2006-09-22 01:56:09 · answer #7 · answered by mathematician 7 · 1⤊ 0⤋
Sphere One
2016-11-01 06:37:11 · answer #8 · answered by Anonymous · 0⤊ 0⤋
THE MAX POSSIBLE LENGTH IS THE SUM OF THE DIAMETERS OF THE CIRCLES=7+7+4+4=22UNITS
2006-09-21 22:47:10 · answer #9 · answered by Anonymous · 0⤊ 0⤋
a million. A= 4pi r^2 dA/dr= 8pi.r =8 pi.2 = 16 pi and dr/dA = a million/(16pi) V= (4/3) pi.r^3 dV/dr = 4pi.r^2 = 4pi.(2^2) = 16pi dV/dA = dV/dr x dr/dA = 16pi . a million/(16pi) =a million dV/dt = dV/dA x dA/dt = a million x 3pi = 3pi cm^3/sec
2016-11-23 14:42:05 · answer #10 · answered by akien 4 · 0⤊ 0⤋